Calc Notes 0304

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(Section 3.4: Derivatives of Trigonometric Functions) 3.4.1

SECTION 3.4: DERIVATIVES OF TRIGONOMETRICFUNCTIONS

LEARNING OBJ ECTIVES

Use the Limit Definition of the Derivative to find the derivatives of thebasic sine and cosine functions. Then, apply differentiation rules to obtainthe derivatives of the other four basic trigonometric functions.

Memorize the derivatives of the six basic trigonometric functions and beable to apply them in conjunction with other differentiation rules.

PART A: CONJ ECTURING THE DERIVATIVE OF THE BASIC SINEFUNCTION

Let f x( ) = sin x . The sine function isperiodicwith period 2 . One cycle of itsgraph is in bold below. Selected [truncated] tangent linesand their slopes(m) areindicated in red. (The leftmost tangent line and slope will be discussed in Part C.)

Remember thatslopesof tangent lines correspond toderivativevalues (that is,values of f ).

The graph of f must then contain the five indicated points below, since their

y-coordinates correspond to values of f .

Do you know of a basic periodic function whose graph contains these points?

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We conjecture that f x( ) = cosx . We will prove this in Parts D and E.

PART B: CONJ ECTURING THE DERIVATIVE OF THE BASIC COSINEFUNCTION

Let g x( ) = cosx . The cosine function is also periodic with period 2 .

The graph of g must then contain the five indicated points below.

Do you know of a (fairly) basic periodic function whose graph contains thesepoints?

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We conjecture that g x( ) = sinx . If f is the sine function from Part A, then we

also believe that f x( ) = g x( ) = sinx . We will prove these in Parts D and E.

PART C: TWO HELPFUL L IMIT STATEMENTS

Helpful Limit Statement #1

limh0

sinh

h= 1

Helpful Limit Statement #2

limh0

cosh 1h

= 0 or, equivalently, limh0

1 coshh

= 0

These limit statements, which are proven in Footnotes 1 and 2, will help usproveour conjectures from Parts A and B. In fact, only thefirst statement is needed forthe proofs in Part E.

Statement #1 helps us graph y =sinx

x.

In Section 2.6, we proved that limx

sinx

x

= 0 by the Sandwich (Squeeze)

Theorem. Also, limx

sinx

x

= 0.

Now, Statement #1 implies that limx0

sin x

x

= 1, where we replacehwithx.

Becausesin x

x

is undefined at x= 0 and limx0

sin x

x

= 1, the graph has ahole

at the point 0,1( ) .

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(Axes are scaled differently.)

Statement #1 also implies that, if f x( ) = sin x , then f 0( ) = 1.

f 0( ) = limh0

f 0 + h( ) f 0( )

h

= limh0

sin 0 + h( ) sin 0( )

h

= limh0

sinh 0

h

= limh0

sinh

h

= 1

This verifies that thetangent lineto the graph of y = sinx at theorigin does,

in fact, haveslope1. Therefore, the tangent line is given by theequationy = x .

By thePrinciple of Local Linearity from Section 3.1, we can say that sin x xwhen x 0 . That is, the tangent line closelyapproximates the sine graph close tothe origin.

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PART D: STANDARD PROOFS OF OUR CONJ ECTURES

Derivatives of the Basic Sine and Cosine Functions

1) Dx

sinx( ) = cosx

2) Dx cosx( ) = sinx

Proof of 1)

Let f x( ) = sin x . Prove that f x( ) = cosx .

f x( ) = limh0

f x+ h( ) f x( )h

=

limh0

sin x+ h( ) sin x( )h

= limh0

sinxcosh+cosxsinh

by Sum Identity for sine

sinxh

= limh0

sinxcosh sinx( )Group terms with sinx.

+cosxsinh

h

= limh0

sinx( ) cosh1( )+cosxsinhh

Now, group expressions containing h.( )

= limh0

sinx( )cosh1

h

0

+ cosx( )sinh

h

1

= cosx

Q.E.D.

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Proof of 2)

Let g x( ) = cosx . Prove that g x( ) = sinx .(This proof parallels the previous proof.)

g x( ) = limh0

g x+ h( ) g x( )h

= limh0

cos x+ h( )cos x( )h

= limh0

cosxcosh sinxsinhby Sum Identity for cosine

cosxh

= limh0

cosxcoshcosx( )

Group terms with cosx.

sinxsinhh

= limh0

cosx( ) cosh1( ) sinxsinhh

Now, group expressions containing h.( )

= limh0

cosx( )cosh1

h

0

sinx( )sinh

h

1

= sinx

Q.E.D.

Do you see where the sign in sin x arose in this proof?

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PART E: MORE ELEGANT PROOFS OF OUR CONJ ECTURES

Derivatives of the Basic Sine and Cosine Functions

1) Dx

sinx( ) = cosx

2) Dx cosx( ) = sinx

Version 2 of the Limit Definition of the Derivative Function in Section 3.2, Part A,provides us with more elegant proofs. In fact, they do not even use Limit Statement#2 in Part C.

Proof of 1)

Let f x( ) = sin x . Prove that f x( ) = cosx .

f x( ) = limh0

f x+ h( ) f x h( )2h

= limh0

sin x+ h( ) sin x h( )2h

= limh0

sinxcosh+cosxsinh( )by Sum Identity for sine

sinxcosh cosxsinh( )by Difference Identity for sine

2h

= limh0

2cosxsinh

2h

= limh0

cosx( )sinh

h

1

= cosx

Q.E.D.

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Proof of 2)

Let g x( ) = cosx . Prove that g x( ) = sinx .

g x( ) = limh0g x+ h( ) g x h( )

2h

= limh0

cos x+ h( )cos x h( )2h

= limh0

cosxcosh sinxsinh( )from Sum Identity for cosine

cosxcosh+ sinxsinh( )from Difference Identity for cosine

2h

= limh0

2sinxsinh

2h

= limh0

sinx( )sinh

h

1

= sinx

Q.E.D.

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A Geometric Approach

Jon Rogawski has recommended a more geometric approach, one thatstresses the concept of the derivative. Examine the figure below.

Observe that: sin x + h( ) sinx h cosx , which demonstrates that thechange in a differentiable function on a small interval h is related to itsderivative. (We will exploit this idea when we discussdifferentials inSection 3.5.)

Consequently,sin x + h( ) sinx

h cosx .

In fact, Dxsinx( ) = lim

h0

sin x + h( ) sin x( )

h= cosx .

A similar argument shows: Dx

cosx( ) = limh0

cos x + h( ) cos x( )

h= sinx .

Some angle and length measures in the figure areapproximate, thoughthey become more accurate as h 0 . (For clarity, the figure does notemploy a small value ofh.)

Exercises in Sections 3.6 and 3.7 will show that thetangent lineto any

pointP on acirclewith center O isperpendicular to the line segmentOP .

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PART F: DERIVATIVES OF THE SIX BASIC TRIGONOMETRIC FUNCTIONS

Basic Trigonometric Rules of Differentiation

1) Dx

sinx( ) = cosx 2) Dx

cosx( ) = sinx

3) Dx

tanx( ) = sec2 x 4) Dx cot x( ) = csc2 x

5) Dx

secx( ) = secx tanx 6) Dx

cscx( ) = cscx cot x

WARNING 1: Radians. We assume thatx, h, etc. are measured inradians(corresponding to real numbers). If they are measured in degrees, the rules of thissection and beyond would have to be modified. (Footnote 3 in Section 3.6 willdiscuss this.)

TIP 1:Memorizing.

The sine and cosine functions are a pair ofcofunctions, as are the tangentand cotangent functions and the secant and cosecant functions.

Lets say you know Rule 5) on the derivative of thesecant function. Youcan quickly modify that rule to find Rule 6) on the derivative of thecosecantfunction.

You take sec xtan x, multiply it by 1 (that is, do a sign flip), and

take thecofunction of each factor. We then obtain: csc xcot x, which is

Dx

cscx( ) .

This method also applies to Rules 1) and 2) and to Rules 3) and 4).

The Exercises in Section 3.6 will demonstrate why this works.

TIP 2:Domains. In Rule 3), observe that tan x and sec2 x share the same domain.In fact, all six rules exhibit the same property.

Rules 1) and 2) can be used toproveRules 3) through 6). The proofs for Rules 4)and 6) are left to the reader in the Exercises for Sections 3.4 and 3.6 (where the

Cofunction Identitieswill be applied).

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(Section 3.4: Derivatives of Trigonometric Functions) 3.4.11 Proof of 3)

Dx

tanx( ) = Dx

sinx

cosx

Quotient Identities( )

= Lo D Hi( ) Hi D Lo( )Lo( )

2Quotient Rule of Differentiation( )

=

cosx Dx sinx( ) sinx Dx cosx( ) cosx( )

2

=

cosx cosx sinx sinx

cosx( )2

=cos2 x+ sin2 x

cos2 xCan: =

cos2 x

cos2 x+

sin2 x

cos2 x=1+ tan2 x= sec2 x

=1

cos2 xPythagorean Identities( )

= sec2 x Reciprocal Identities( )

Q.E.D.

Footnote 3 gives a proof using the Limit Definition of the Derivative.

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Proof of 5)

Dx

secx( ) = Dx

1

cosx

Quotient Rule of Differentiation Reciprocal Rule

=

LoD Hi( ) Hi D Lo( )

Lo( )2

or D Lo( )

Lo( )2

=

cosx Dx 1( ) 1 Dx cosx( ) cosx( )

2or

Dx

cosx( )

cosx( )2

=

cosx 0 1 sinx

cosx( )2 or

sinx

cosx( )2

=

sinx

cos2 x

=

1

cosxsinx

cosxFactoring or "Peeling"( )

= secxtanx Reciprocal and Quotient Identities( )

Q.E.D.

Example 1 (Finding a Derivative Using Several Rules)

Find Dx

x2secx + 3cosx( ).

Solution

We apply theProduct Rule of Differentiation to the first term and theConstant Multiple Ruleto the second term. (The Product Rule can be usedfor the second term, but it is inefficient.)

Dx

x2

secx + 3cosx( ) = Dx x2 secx( )+ Dx 3cosx( ) Sum Rule of Diff'n( )

= Dx

x2( ) secx[ ]+ x

2 Dx secx( ) ( )+ 3 Dx cosx( ) = 2x[ ] secx[ ]+ x2 secx tanx[ ]( ) + 3 sinx[ ]= 2x secx + x

2secx tanx 3sinx

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Example 2 (Finding and Simplifying a Derivative)

Let g ( ) =cos

1 sin. Find g ( ) .

Solution

Note: Ifg ( ) werecos

1 sin2

, we would be able to simplify considerably

before we differentiate. Alas, we cannot here. Observe that we cannot splitthe fraction through its denominator.

g ( ) =Lo D Hi( ) Hi D Lo( )

Lo( )2

Quotient Rule of Diff'n( )

=1 sin[ ] D cos( ) cos[ ] D 1 sin( )

1 sin( )2

Note: 1 sin( )2

isnot equivalent to 1 sin2 .

=

1 sin[ ] sin[ ] cos[ ] cos[ ]1 sin( )

2

TIP 3:Signs. Many students dont see why

Dsin( ) = cos. Remember that differentiating the

basicsinefunction doesnot lead to a sign flip, whiledifferentiating the basiccosinefunctiondoes.

= sin+ sin

2+ cos

2

1 sin( )2

WARNING 2: Simplify.

=

sin+

11 sin( )

2 Pythagorean Identities( )

=1 sin

1 sin( )2

Rewriting( )

=1

1 sin

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Example 3 (Simplifying Before Differentiating)

Let f x( ) = sin x csc x . Find f x( ) .

Solution

Simplifying f x( ) first is preferable to applying the Product Rule directly.

f x( ) = sinx cscx

= sinx( )1

sinx

Reciprocal Identities( )

= 1, sinx 0( )

f x( ) = 0, sin x 0( )

TIP 4:Domain issues.

Dom f( ) = Dom f( ) = x sinx 0{ } = x x n, n ( ){ } .In routine differentiation exercises, domain issues are often ignored.

Restrictions such as sin x 0( ) here are rarely written.

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PART G: TANGENT L INES

Example 4 (Finding Horizontal Tangent Lines to a Trigonometric Graph)

Let f x( ) = 2sin x x . Find thex-coordinates of all points on the graph of

y=

f x( ) where thetangent lineishorizontal. Solution

We must find where theslopeof the tangent line to the graph is 0.We must solve the equation:

f x( ) = 0

Dx 2sinx x( ) = 0

2cosx 1 = 0

cosx = 12

x =

3+ 2n, n ( )

The desiredx-coordinates are given by:

x x =

3+ 2n, n ( )

.

Observe that there areinfinitely manypoints on the graph where the

tangent line is horizontal.

Why does the graph of y = 2sinx x below make sense? Observe that f is

anodd function. Also, the x term leads to downward drift; the graph

oscillates about the line y = x .

Thered tangent linesbelow are truncated.

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Example 5 (Equation of a Tangent Line; Revisiting Example 4)

Let f x( ) = 2sin x x , as in Example 4. Find an equation of thetangent line

to the graph ofy = f x( ) at the point

6, f

6

.

Solution

f

6

= 2sin

6

6= 2

1

2

6= 1

6, so thepoint is at

6, 1

6

.

Findm, theslopeof the tangent line there. This is given by f

6

.

m = f

6

Now, f x( ) = 2cosx 1 (see Example 4).

= 2cos

6

1

= 23

2

1

= 3 1

Find aPoint-Slope Formfor the equation of the desired tangent line.

y y1= m x x

1( )

y 1

6

= 3 1( ) x

6

, or y 6

6= 3 1( ) x

6

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FOOTNOTES

1. Proof of L imit Statement #1 in Part C. First prove that lim0+

sin

= 1, where we use to

represent angle measures instead ofh.

Side note: The area of a circular sector such asPOB below is given by:

Ratio of to a

full revolution

Area of

the circle

=

2

r2( ) =

1

2r

2=

1

2 ifr = 1( ) , where 0, 2[ ] .

Area of TrianglePOA Area of SectorPOB Area of TriangleQOB

1

2sincos

1

2

1

2tan, 0,

2

sincos tan, " "

cos

sin

1

cos, " "

Now, we take reciprocals and reverse the inequality symbols.( )

1

cos

sin

cos, " "

cos

1

sin

1

cos1

, 0,

2

as 0+

. Therefore, lim0+sin

= 1 by a one-sided variation of the

Squeeze (Sandwich) Theorem from Section 2.6.

Now, prove that lim0

sin

= 1. Let = .

lim0

sin

= lim

0+

sin ( )

= lim0+

sin ( )

= lim0+

sin ( )

= 1. Therefore, lim0

sin

= 1.

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(Section 3.4: Derivatives of Trigonometric Functions) 3.4.18.

2. Proof of Limit Statement #2 in Part C.limh0

cosh 1h

= limh0

1 cosh

h

= lim

h0

1 cosh( )h

1+ cosh( )1+ cosh( )

= lim

h0

1 cos2 hh 1+ cosh( )

= limh0

sin2h

h 1+ cosh( )

= lim

h0

sinh

h

sinh

1+ cosh

=

limh0

sinh

h

limh0

sinh

1+ cosh

= 1( ) 0( ) = 0 .

3. Proof of Rule 3) Dx

tanx( ) = sec2x, using the Limit Definition of the Derivative.

Let f x( ) = tanx . f x( ) = limh0

f x + h( ) f x( )h

= limh0

tan x + h( ) tan x( )h

= limh0

tanx + tanh

1 tanx tanh tanx

h

= limh0

tanx + tanh

1 tanx tanh tanx

h

1 tanx tanh( )

1 tanx tanh( )

= limh0

tanx + tanh tanx( ) 1 tanx tanh( )h 1 tanx tanh( )

= limh0

tanh + tan2x tanh

h 1 tanx tanh( )

= limh0

tanh( ) 1+ tan2 x( )h 1 tanx tanh( )

= limh0

tanh( ) sec2 x( )h 1 tanx tanh( )

(Assume the limits in the next step exist.)

= limh

0

tanh

h

sec

2x( ) lim

h

0

1

1 tanx tanh

= limh

0

sinh

cosh

1

h

sec

2x( ) 1( )

= limh0

sinh

h

1

cosh

sec

2x( ) = lim

h0

sinh

h

lim

h0

1

cosh

sec2 x( ) = 1( ) 1( ) sec2 x( )

= sec2x

4. A joke.The following is a sin: sin xx

= sin . Of course, this is ridiculous!

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Calc Notes 0304